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inertia formula for con rod

I am trying to calculate all the forces (in lbf) that the connecting rod adds to the engine which influences vibration.
I am dividing the rod in between the two ends into four sections and calculating for the middle point of each as if that point had all that sections weight.
I am calculating for every 15 degrees of crank rotation.conrodforces.gif
Hereís the formulas I have but Iím not sure about the in-line force formula:
centrifugal force in lbf = (kg x (meters/sec velocity)^2 / arc radius in meters) x .225
in-line force in lbf = (kg x (meters/sec beginning velocity^2) / 2) x .74

I am smudging a bit by ignoring the curvature of movement for calculating the in-line force but I donít care if itís 10% off.
I think ďinertia forceĒ is the same as what Iím calling in-line force because to stop a moving object you have to exert a force equal to its inertia. Correct me if Iím wrong.

In most ancient texts, the connecting rod mass is divided into two parts at the center of connecting rod mass. The big end of the connecting rod around the crank pin was considered to be a rotating mass. The small end of the connecting rod was considered to have only linear motion with the piston. The lumped masses of piston and pin assembly along with mass of the connecting rod 'above' the center of mass is used to calculate the fluctuating reciprocating force. This is considered to be linear or in-line as you call it parallel to the bore axis. This is an approximation but one that is commonly used.

I am not familiar with a connecting rod force approximation that uses four parts.